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Determinants

Short Answer Type

311.

For what values of k, the system of linear equations

x + y + z = 2
2x + y - z = 3
3x + 2y + kz = 4

has a unique solution?

1308 Views

Long Answer Type

312.

Using properties of determinants, prove that

$open vertical bar table row cell left parenthesis straight x plus straight y right parenthesis squared end cell zx zy row zx cell left parenthesis straight z plus straight y right parenthesis squared end cell xy row zy xy cell left parenthesis straight z plus straight x right parenthesis squared end cell end table close vertical bar space space equals space 2 xyz left parenthesis straight x plus straight y plus straight z right parenthesis cubed$

1198 Views

Short Answer Type

313.
Using the properties of determinants, solve the following for x:

$open vertical bar table row cell straight x plus 2 end cell cell space space straight x plus 6 end cell cell space straight x minus 1 end cell row cell straight x plus 6 end cell cell straight x minus 1 end cell cell straight x plus 2 end cell row cell straight x minus 1 end cell cell straight x plus 2 end cell cell straight x plus 6 end cell end table close vertical bar space equals space 0$

Let space increment space equals space open vertical bar table row cell straight x plus 2 end cell cell space space straight x plus 6 end cell cell space straight x minus 1 end cell row cell straight x plus 6 end cell cell straight x minus 1 end cell cell straight x plus 2 end cell row cell straight x minus 1 end cell cell straight x plus 2 end cell cell straight x plus 6 end cell end table close vertical bar space equals space 0

Applying space straight C subscript 2 rightwards arrow straight C subscript 2 minus straight C subscript 1 space and space straight C subscript 3 space rightwards arrow space straight C subscript 3 space minus straight C subscript 1 increment space equals space open vertical bar table row cell straight x plus 2 end cell cell space space 4 end cell cell space space minus 3 end cell row cell straight x plus 6 end cell cell space minus 7 end cell cell space minus 4 end cell row cell straight x minus 1 end cell cell space 3 end cell cell space space space 7 end cell end table close vertical bar Applying space straight R subscript 2 space rightwards arrow straight R subscript 2 minus straight R subscript 1 space and space straight R subscript 3 space rightwards arrow straight R subscript 3 minus straight R subscript 1 space space space increment equals space open vertical bar table row cell straight x plus 2 end cell cell space space 4 end cell cell space space minus 3 end cell row 4 cell space minus 11 end cell cell space minus 1 end cell row cell negative 3 end cell cell space minus 1 end cell cell space space 10 end cell end table close vertical bar space Applying space straight R subscript 2 rightwards arrow straight R subscript 2 space plus space straight R subscript 3 increment space equals space open vertical bar table row cell straight x plus 2 end cell cell space space 4 end cell cell space space minus 3 end cell row 1 cell space minus 12 end cell cell space space space 9 end cell row cell negative 3 end cell cell space minus 1 end cell cell space 10 end cell end table close vertical bar Applying space straight R subscript 3 space rightwards arrow straight R subscript 3 plus left parenthesis 3 right parenthesis straight R subscript 2 space increment space equals space open vertical bar table row cell straight x plus 2 end cell cell space 4 end cell cell space minus 3 end cell row 1 cell negative 12 end cell cell space space 9 end cell row 0 cell negative 37 end cell cell space 37 end cell end table close vertical bar Expanding space along space straight C subscript 1

increment space equals space left parenthesis straight x plus 2 right parenthesis space open vertical bar table row cell negative 12 end cell cell space space 9 end cell row cell negative 37 end cell cell space 37 end cell end table close vertical bar minus 1 open vertical bar table row 4 cell space minus 3 end cell row cell negative 37 end cell cell space space 37 end cell end table close vertical bar increment equals left parenthesis straight x plus 2 right parenthesis thin space left parenthesis negative 444 plus 333 right parenthesis minus 1 left parenthesis 148 minus 111 right parenthesis increment equals space left parenthesis straight x plus 2 right parenthesis thin space left parenthesis negative 111 right parenthesis minus 1 left parenthesis 37 right parenthesis therefore space increment space equals space 0 space equals space minus 111 straight x minus 259 therefore space space straight x space equals space minus 259 over 111 equals negative 7 over 3

751 Views

314.

If $open vertical bar table row cell 3 straight x end cell cell space 7 end cell row cell negative 2 end cell cell space 4 end cell end table close vertical bar space equals open vertical bar table row 8 cell space 7 end cell row 6 cell space 4 end cell end table close vertical bar comma$ find the value of x.

196 Views

315.

Use Properties of determinants, prove that:
$open vertical bar table row cell 1 plus straight a end cell cell space 1 end cell cell space 1 end cell row 1 cell 1 plus straight b end cell 1 row 1 1 cell 1 plus straight c end cell end table close vertical bar space equals space abc plus bc plus ca plus ab$

389 Views

316.

If $open vertical bar table row cell straight x plus 1 end cell cell space space straight x minus 1 end cell row cell straight x minus 3 end cell cell space straight x plus 2 end cell end table close vertical bar space equals space open vertical bar table row 4 cell space space minus 1 end cell row 1 cell space space space space 3 end cell end table close vertical bar comma$ then write the value of x.

193 Views

317.

Using properties of determinants prove the following:
$open vertical bar table row 1 cell space space straight x end cell cell space space straight x squared end cell row cell straight x squared end cell cell space 1 end cell straight x row straight x cell space straight x end cell 1 end table close vertical bar space equals space left parenthesis 1 minus straight x cubed right parenthesis squared$

247 Views

318.

Using properties of determinants, prove that

594 Views

Long Answer Type

319.

Use product $open square brackets table row 1 cell negative 1 end cell 2 row 0 2 cell negative 3 end cell row 3 cell negative 2 end cell 4 end table close square brackets space space open square brackets table row cell negative 2 end cell 0 1 row 9 2 cell negative 3 end cell row 6 1 cell negative 2 end cell end table close square brackets$ to solve the system of equations x + 3z = 9,
–x + 2y – 2z = 4, 2x – 3y + 4z = –3